Buchmann- Meyer Light patterns

[subkontrabob]

Hi all,

I recently got interested in Buchmann Meyer patterns, and I've read the Axon/Geddes AES paper so far.

Can anyone recommend further reading? Googling and the reference lists of the few articles I have only turn up 50 - 80 year old articles published in long defunct magazines impossible to get hold of.

The original paper by Buchmann and Meyer was published in 1930!

Does anyone by chance have some of this old stuffed scanned?





best regards,

Bob

[Kris D]

The Oliver Read book has some info on cutting Meyer buckman patterns. Make sure you bypass RIAA when making the tests!!

Kris

[subkontrabob]

Thanks a million Kris! The part about the light patterns is "hidden" in the audio measurements chapter in the end of the book - I haven't read the entire book yet, just looked at the disk cutting chapters, and totally missed that part!

[markrob]

Hi,

I have the original paper by them ca. 1940. If you PM me with your email, I'll scan and send it. Do you have the AES Disc Recording Anthologies? If so there is a great paper titled: "The Calibration of Disc Recordings by Light-Pattern Measurments" by P.E. Axon, W.K.E. Geddes. Its a bit too long to scan, but goes into great detail on how to make quantitative measurments using this method. If you don't have this collection of papers, I'd highly reccomend it.

Mark

[subkontrabob]

Hi Mark,

thank you for the kind offer! I've sent you a PM. I have both volumes of the anthology, and I've read the article by Axon and Geddes.

By the way, does anyone use this method currently?

[cohearent]

Although non-RIAA is the easiest way to actually measure the response, by sweeping with a programmable sweep oscillator you can sure see any "bumps" in response easily

[subkontrabob]

Although non-RIAA is the easiest way to actually measure the response, by sweeping with a programmable sweep oscillator you can sure see any "bumps" in response easily

Hmmmm, has anyone tried to program a sine sweep that would give equally spaced octaves on a cut with a constant given LPI?

[greybeard]

@subkontrabob
what you are describing looks like a logarithmic sweep, which is standard. However, I cannot see how it could be affected by the LPI. But perhaps I am just not understanding the pattern you would like to see.
Best!

[subkontrabob]

Hey greybeard,

you are right, a logarithmic sweep would be the start. The thing I was wondering about is, how much compensation the constantly shrinking circumference of the groove would need, or if it is a neglectable factor.

[greybeard]

Hey Subcontrabob,

just for anybody out there who has not recently read details about the Buchmann-Meyer “tree” or light-bandwidth (width of the light band), I would just repeat the simple relationship. Recorded velocity in cm/s is equal to the light bandwidth in mm times the rpm of the record, divided by 60pi.

Or, if you want the light bandwidth in mm: it is equal to 60pi times the velocity in cm/s divided by the rpm of the record. (60pi is about 190)

Hence it is completely independent of the declining linear velocity as you get nearer the centre.

What you observe when you look at a groove through the microscope is the peak-to-peak amplitude, and it is this that is limited by the groove pitch in lines per inch. When you look at a record via a Buchmann-Meyer setup you are observing the recorded velocity for a given radius. What you are actually seeing is the reflection from the relatively “flat” passages midway between the amplitude peaks, i.e. the zero crossings. This is the maximum of the first derivative, mathematically, of the amplitude function.

If we look at sinusoidal signals only, there is a simple relationship between the maximum velocity and the maximum amplitude there is room for relating to a given groove pitch: if you record an upwards sweep with constant maximum amplitude, your velocity will increase proportionally with frequency, and the zero crossings (this is where the maximum steepness is) will be impossible to trace. Also, if you try to cut very steep zero crossings, the back of the stylus will destroy what you just cut with the face of the stylus.

For these reasons and because of the fact that the best pickups are velocity pickups (output proportional to the velocity of a change of magnetic field rather than excursion or deflection from a midpoint=zero), records are essentially recorded with constant maximum velocity, ensuring that the zero crossings never go above a certain limit in steepness. However, this way the proportionality is inverse for the amplitude: it decreases according to inverse proportion to the frequency. This would mean that if you cut only a sinusoidal signal outside in, you could increase the groove pitch in LPI proportional to the frequency and get more room for lovely sinusoidal signal. If you do a logarithmic sweep the total time needed for a sweep from 20 Hz to 20 kHz would be much reduced – it is only 3 decades. So, using a logarithmic sweep and automatic control of LPI you could record really long sweeps.

However, the decrease of amplitude as we go higher in frequency also means that amplitude increases if we go lower in frequency, and it quickly becomes too large for the LPI. For this reason, from a certain “knee”, “crossover”, “turnover”, or whatever you want to call it, the recording characteristic changes to constant amplitude when we go down in frequency. The good velocity pickup mentioned above will give lower and lower signals, and we have to increase the amplification in the preamp as we go lower: the reproduction characteristic, that is ideally the inverse of the recording characteristic.

We have not at all discussed the acceleration of the cutting stylus, which is why we need such high power to drive it when we use feedback, nor the acceleration of the tip of the reproducing stylus, which gives rise to wear on the record. Suffice it to say here that that is again the derivative of the velocity, i.e. the second derivative of the amplitude. Good cutting computers keep track of all three and adjust and protect accordingly.

Best!

[subkontrabob]

Hi greybeard,

I think you misinterpreted my post. I wasn't talking about the light band width, I understand the underlying principle. I was talking about the distribution of the different frequencies along the groove if you cut a one side-long logarithmic sine sweep on a disc. To get equal octave distances you would have to compensate for the decreasing groove circumference.

The sweep speed would have to start slow and then be slowly accelerated.

It would probably make most sense to start at the high frequencies to avoid the high frequency loss in the innermost grooves.

[greybeard]

Hey Subcontrabob,

you think I misinterpreted your post; well yes and no.

I wasn't talking about the light band width, I understand the underlying principle.

… well, strictly I was not writing to you – you had read Axon and Geddes – but to anybody who was not conversant.

I was talking about the distribution of the different frequencies along the groove if you cut a one side-long logarithmic sine sweep on a disc.

… so far, so good, I follow you

To get equal octave distances you would have to compensate for the decreasing groove circumference.

… this is where I lose you. Which distances are you talking about? What is the usefulness of equal octave distances? If you take one frequency as the start, the octaves in a sweep are only present at precisely 2, 4, 8, … times that frequency.

The sweep speed would have to start slow and then be slowly accelerated.

… well, that is the logarithmic sweep with time as the independent axis. So you would get an octave at constant intervals of time.

It would probably make most sense to start at the high frequencies to avoid the high frequency loss in the innermost grooves.

… but then you would want to do an inverse logarithmic sweep. Anyway, most of the 78 rpm test records I know are made this way, except Polydor/Deutsche Grammophon/Brunswick, which go up to 10 kHz at the small radius.

I may definitely be too dense, but I have to admit that I do not understand the problem.

Best!

[Phinster]

The NAB test record has a 'christmas tree' sweep on it.

Someone told me turntable speed was varied to cut the HF.

Don't see how they could accomplish this, with only one groove on the side.

Probably nonsense!